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Let $G$ be a finite group and $p$ be a prime divisor of $|G|$. Consider the set $X=\{g\in G:g^p=1\}$. Show that $p$ divides $|X|$.

My attempt: Consider the action of $G$ on $X$ by conjugation. Then ${\rm Stab}_g=C_G(g)$ for all $g\in X$. I'm stuck here.

Shaun
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Guest
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    Look at this https://math.stackexchange.com/a/238424/1070376 – calc ll Jul 13 '22 at 23:52
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    So this looks like a rather difficult problem. And the result is attributed to McKay. – calc ll Jul 14 '22 at 04:04
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    Just a remark. Frobenius has proved that if $n$ divides the order of a group $G$, then the number of elements whose $n$-th power is $1$ is a multiple of $n$. – Andreas Caranti Jul 15 '22 at 10:54
  • There is a nice proof of Frobenius' theorem here https://www.jstor.org/stable/2324902#metadata_info_tab_contents – Andreas Caranti Jul 15 '22 at 10:58
  • @Cpc: Would you mind making a short answer out of your comment? (This question is not a strict duplicate of https://math.stackexchange.com/q/238385/61691, so we should make sure that it does not stay in open but unanswered state.) – azimut Jul 16 '22 at 08:57
  • Well, @azimut, the problem is solved in the answer linked above. What should I say? – calc ll Jul 16 '22 at 09:01
  • @Cpc it completely agree. But technically, a comment is not an answer in the stackexchange mechanism, so this question counts as unanswered. So it would be great if you could formulate your comment as a short "real" answer. – azimut Jul 16 '22 at 13:20

1 Answers1

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$X$ comprises all and only the elements of $G$ of order $p$, plus the identity. Such elements are grouped in $m$ trivially intersecting subgroups each of order $p$. Therefore: \begin{alignat}{1} |X| &= m(p-1)+1 \\ &=mp-(m-1) \\ \tag1 \end{alignat} But $m\equiv 1\pmod p$, because the number $n_k$ of subgroups of $G$ of order $p^k$ is congruent to $1$ modulo $p$ for every $k=0,1,\dots,k_{\text{max}}$ (see e.g. here), and in particular for $k=1$. Therefore, $m=n_1\equiv 1\pmod p$, and hence from $(1)$ $p\mid |X|$.

  • What does the phrase "arraied in $m$, trivially intersecting subgroups" mean? – Guest Jul 18 '22 at 19:17
  • By Lagrange, two subgroups of order $p$ either coincide or trivially intersect, @Guest. And all the nontrivial elements of a subgroup of order $p$ have order $p$. –  Jul 18 '22 at 19:18
  • Yes, I know. But what is the meaning of "arraied"? – Guest Jul 18 '22 at 19:21
  • I meant "are placed in". –  Jul 18 '22 at 19:23
  • So these subgroups are placed in an integer? – Guest Jul 18 '22 at 19:25
  • Think of a flower: each petal is a subgroup of order $p$, and each petal touch any other in center of the flower (the identity), only. Each petal comprises $p-1$ elements (each of order $p$), plus the identity; and there are $m$ petals. –  Jul 18 '22 at 19:29
  • Oh, the comma after $m$ is what caused the entire confusion! – Guest Jul 18 '22 at 19:52
  • Oops, sorry, you are right it may be confusing... I'm used in that when there are multiple adjectives referring to one same noun. –  Jul 18 '22 at 19:56
  • Comma removed now. –  Jul 18 '22 at 19:57
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    I corrected the spelling of 'arrayed', which might have been throwing some people off. – David A. Craven Jul 19 '22 at 07:42
  • Thank you, @DavidA.Craven. Thanks to MSE, I'm learning probably as much English as math! –  Jul 19 '22 at 07:45
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    @fitzcarraldo I maybe would use 'grouped' because 'arrayed' usually would mean set out in rows and columns, like a matrix, a table, or an old-fashioned army. – David A. Craven Jul 19 '22 at 07:47
  • Good, edited accordingly. –  Jul 19 '22 at 07:49